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Let $x$ be a positive real number. Then

  1. $x^2+\pi ^2 + x^{2 \pi} > x \pi+ (\pi + x) x^{\pi}$
  2. $x^{\pi}+\pi^x > x^{2 \pi} + \pi ^{2x}$
  3. $\pi x +(\pi+x)x^{\pi} > x^2+\pi ^2 + x^{2 \pi}$
  4. none of the above
in Numerical Ability by Veteran (431k points)
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1 Answer

+1 vote

Here, $x$ is a positive real number i.e. $x>0$. 

If $x=1$, the option B is false as $1^\pi +\pi^1=1+\pi$ and $1^{2\pi} +\pi^2=1+\pi^2$. Definitely $1+\pi<1+\pi^2$.

 

Now let's check for the option A and C.

When $0<x<1$

It means $x$ is a fraction (floating point numbers).
Now $\pi>3\Leftrightarrow \pi^2>9$

Again $\pi<4\Leftrightarrow 2\pi<8\Leftrightarrow2\pi+1<9 $

Combining them yields $2\pi+1<\pi^2 \tag{i}$

Now

$\begin{align}x&< 1 \\ \Rightarrow x^{\pi}&< 1 \tag{ii}\end{align}$

$\begin{align}\therefore x+x^{\pi}&< 2 \\ \Rightarrow (x+x^{\pi})\pi&< 2\pi \end{align} \tag{iii}$

Now from no$\mathrm{(ii)}$ multiplying both sides by $x$, we get

$\begin{align} x\cdot x^\pi &< 1\cdot x \\ \Rightarrow x^{\pi+1} &< x \tag{iv} \end{align}$

no$\mathrm{(iii)}+$no$\mathrm{(iv)}\Rightarrow$

$\begin{align}(x+x^{\pi})\pi+x^{\pi+1} &< 2\pi+ x \\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< 2\pi+x \\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< 2\pi+1; ~[\because x<1 \Rightarrow 2\pi+x< 2\pi+1]\\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< \pi^2; ~[\mathrm{From~no(i)}] \end{align}$

 

As $x>0$, definitely $\pi^2<x^2+\pi^2+x^{2\pi}$.

$$\therefore x\pi+(\pi+x)x^{\pi} < x^2+\pi^2+x^{2\pi}$$

So, the option C is false. Thus the option A is true.

When $x\ge1$

Let $g(x)=x^{2\pi}+x^2+\pi^2$, $h(x)=x\pi+(\pi+x)x^{\pi}=x^{\pi+1}+\pi x^{\pi}+\pi x$ and $f(x)=g(x)-h(x)$.

$\begin{align}\therefore f'(x)&=g'(x)-h'(x)\\ &=2\pi x^{2\pi-1}+2x-(\pi+1)x^\pi-\pi^2 x^{\pi-1}-\pi\end{align}$

Now, $f'(1)<0$ and $f'(2)>0$. It means $f(x)$ is concave upward (having a minimum value) and there is a root $a \in [1,2]$ such that $f'(a)=0$.  Using bisection method, we can get closer to this root. Then

$f'(\frac{1+2}{2})=f'(1.5)>0$

Then

$f'(\frac{1+1.5}{2})=f'(1.25)<0$

Then

$f'(\frac{1.25+1.5}{2})=f'(1.375)\approx 0$.

So $f(1.375)$ is approximately minimum $=2.553>0$. As the function is concave upward and its minimum value is positive, it means $f(x)>0~\forall~x\ge1$.

$\begin{align} \therefore f(x)&>0 \\ \Rightarrow g(x)-h(x) &>0 \\ \Rightarrow g(x)&>h(x) \\ \Rightarrow x^2+\pi^2+x^{2\pi} &> x\pi+(\pi+x)x^{\pi} \end{align}$

Therefore, $x^2+\pi^2+x^{2\pi} > x\pi+(\pi+x)x^{\pi}$ for all $x>0$.

So the correct answer is A.

by Active (3.6k points)
edited by
+1

Here's the graph of $f(x)=x^{2\pi}+x^{2}+\pi^{2}-\left(x^{\pi+1}+\pi x^{\pi}+\pi x\right)$ below.

 

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